# Double integral evaluation

1. Apr 9, 2005

### stunner5000pt

More fun yaaay
evaluate $$\int \int \int_{G} x^2 yz dx dy dz$$
where G is bounded by plane z=0, z=x, y=1, y=x

certrainly zi s bounded below by 0 and above by x. and y is boundedbelow by 1 and above by x. having a hard time picturing this...

i dont think this would pictured how the double integrals were, is there a way to visualize this?? Let x come out of the apge, y go right and z upwards
then z=0 is the x y plane, y =1 come out the page y=x come out of the and goes right and finally z=x come out the page and goes upward
but i dont think this helped determining the bounds did it??
Plase help

2. Apr 9, 2005

### xanthym

SOLUTION HINTS:
Try the following:

$$1: \ \ \ \ \int_{x=0}^{1} \int_{y=1}^{x} \int_{z=0}^{x} x^2 \cdot y \cdot z \ dz \, dy \, dx \ \ =$$

$$2: \ \ \ \ = \ \ \int_{x=0}^{1} \ x^2 \ \left ( \int_{y=1}^{x} y \ \left( \int_{z=0}^{x} z \, dz \right) \ dy \right ) \ dx \ \ =$$

$$3: \ \ \ \ = \ \ \int_{x=0}^{1} \ x^2 \ \left ( \int_{y=1}^{x} y \ \left( \, \left[ z^{2}/2 \right]_{0}^{x} \, \right) \ dy \right ) \ dx \ \ =$$

$$4: \ \ \ \ = \ \ \int_{x=0}^{1} \ x^2 \ \left ( \int_{y=1}^{x} y \cdot (x^{2}/2) \ dy \right ) \ dx \ \ =$$

$$5: \ \ \ \ = \ \ \int_{x=0}^{1} \ (x^{4}/2) \ \left ( \int_{y=1}^{x} y \ dy \right ) \ dx \ \ =$$

Continue this process until all integrations have been performed and evaluated.

~~

Last edited: Apr 9, 2005
3. Apr 11, 2005

### ehild

Xanthym, the result is negative, that should not be. Change over the bound of integration with respect to y.

Stunner, you should sketch a diagram to find the bounds of integration. You have to find a closed volume as the domain of integration. That is a tetrahedron in this case, with vertexes (0,0,0), (1,1,1), (0,1,0) and (1,1,0) Look at the picture. So it is either

$$\int_{x=0}^{1} \int_{y=x}^{1} \int_{z=0}^{x} x^2 \cdot y \cdot z \ dz \, dy \, dx$$

or

$$\int_{y=0}^{1} \int_{x=0}^{y} \int_{z=0}^{x} x^2 \cdot y \cdot z \ dz \, dx \, dy$$.

ehild

Last edited: Jun 29, 2010
4. Apr 11, 2005

### xanthym

(Bounds on y-integration corrected, now {y="x→1"} instead of previous {y="1→x"}, thanks to ehild's comment.)
SOLUTION HINTS:

$$1: \ \ \ \ \int_{x=0}^{1} \int_{y=x}^{1} \int_{z=0}^{x} x^2 \cdot y \cdot z \ dz \, dy \, dx \ \ =$$

$$2: \ \ \ \ = \ \ \int_{x=0}^{1} \ x^2 \ \left ( \int_{y=x}^{1} y \ \left( \int_{z=0}^{x} z \, dz \right) \ dy \right ) \ dx \ \ =$$

$$3: \ \ \ \ = \ \ \int_{x=0}^{1} \ x^2 \ \left ( \int_{y=x}^{1} y \ \left( \, \left[ z^{2}/2 \right]_{0}^{x} \, \right) \ dy \right ) \ dx \ \ =$$

$$4: \ \ \ \ = \ \ \int_{x=0}^{1} \ x^2 \ \left ( \int_{y=x}^{1} y \cdot (x^{2}/2) \ dy \right ) \ dx \ \ =$$

$$5: \ \ \ \ = \ \ \int_{x=0}^{1} \ (x^{4}/2) \ \left ( \int_{y=x}^{1} y \ dy \right ) \ dx \ \ =$$

$$6: \ \ \ \ = \ \ \int_{x=0}^{1} \ (x^{4}/2) \ \left( \, \left[ y^{2}/2 \right]_{x}^{1} \, \right) \ dx \ \ =$$

Continue this process until all integrations have been performed and evaluated.

~~

Last edited: Apr 11, 2005